Chapter 9�Models of Chemical Bonding by z28wV610

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									Chapter 9—Models of Chemical
          Bonding




         AP Chemistry
    What is a chemical bond?
Forces that cause a group of atoms to
 behave as a unit
       Why do atoms bond?
To lower the potential energy of their atoms
   which creates a more stable
   arrangement of atoms
Figure 9.1

  A general comparison of metals and
              nonmetals
             Types of Chemical Bonding


1. Metal with nonmetal:

       electron transfer and ionic bonding

2. Nonmetal with nonmetal:

       electron sharing and covalent bonding

3. Metal with metal:

       electron pooling and metallic bonding
Figure 9.2
        The three models of chemical bonding
                    Ionic Bond
An atom that loses electron(s) combines
 with an atom that gains electron(s)

Ex: Draw electron configurations for…
Na                        Cl
              1
1s 22s22p6          1s              5
                       22s22p63s23p 6
           3s

       electrostatic attraction
Na+                               Cl-
             Covalent Bond
• Sharing of electron pairs between two
  atoms
• Shared electrons are simultaneously
  attracted to both atomic nuclei



Video
    Structure of a metallic crystal




The “sea” of delocalized electrons allows atoms to move past one
another as the solid deforms, creating a malleable substance.
Gilbert Newton Lewis (1875-1946)
    •Determined electrode potentials,
    conductivity, free energy and other
    thermodynamic constants for elements.
    •Provided the first description of covalent
    bonding and a shared pair of electrons
    between atoms. Theorized that in most
    atoms electrons arranged themselves so that
    there were 8 electrons around the atoms.
    •First to prepare pure deuterium; predicted
    presence of heavy water.


    Lewis's speculations on
    atomic structure.
                  Lewis Electron-Dot Symbols

  For main group elements -

      The A group number gives the number of valence electrons.
      Place one dot per valence electron on each of the four
      sides of the element symbol.
      Pair the dots (electrons) until all of the valence electrons are
      used.

Example:
           Nitrogen, N, is in Group 5A and therefore has 5 valence
           electrons.
                  .                       .              .
                              :




                : N .        . N.
                               .
                                        . N:           . N.
                  .                       .



                                                        :
             Dot Diagrams
• Symbolize outer shell electrons
         (s and p orbitals only)

• Why?
    Those are the electrons involved in
    bonding and chemical reactions

LEFT RIGHT TOP BOTTOM
                     X
Figure 9.3


    Lewis electron-dot symbols for elements in Periods 2 and 3
     SAMPLE PROBLEM 9.1                        Depicting Ion Formation

     PROBLEM:     Use partial orbital diagrams and Lewis symbols to depict the
                  formation of Na+ and O2- ions from the atoms, and determine
                  the formula of the compound.
     PLAN:      Draw orbital diagrams for the atoms and then move electrons to
                make filled outer levels. It can be seen that 2 sodiums are
                needed for each oxygen.

     SOLUTION:

                                                          O2-
Na
                                                                2s          2p
       3s         3p    O

                             2s          2p                     2 Na+
Na
                                                    Na

                                                      .

                                                            .




                                                                                 :
       3s          3p                                     + : O:        2Na+ + : O :2-




                                                                                 :
                                                                .
                                                    Na.
                  Octet Rule
• When atoms bond,
  they lose, gain, or
  share electrons to
  attain a filled outer
  level of eight (or two)
  electrons
              Ionic Bonding
• Transfer of electrons from metal atoms to
  nonmetal atoms to form ions that come
  together in a solid ionic compound
Figure 9.4
              Three ways to represent the formation of Na+ and F-
                          through electron transfer.
    Electron configurations

Na 1s22s22p63s1 + F 1s22s22p5               Na+ 1s22s22p6      +     F- 1s22s22p6

    Orbital diagrams
                                                   Na+
       Na
                                                         3s             3p
              3s        3p

        + F                                   +    F-
                                                         1s        2s        2p
                   1s       2s         2p

    Lewis electron-dot symbols
                                  .




                                                              :
              Na .      +        :F:         Na+    +       : F: -




                                                              :
                                 :
               Bond Energy
Energy required to break a bond (kJ/mol) (+)
Energy released when a bond is formed (-)
     Calculate DE In Forming NaF
                                                         4

12: Na(s)Na(g)        DE = +109 kJ       3

                                                                  5
                   energy of sublimation
                                           2
                                           1

23: Na(g) Na+ (g) DE = +495 kJ
                                               Overall
                   Ionization Energy           Energy
                                               Change
                                                                  6



34: ½ F2 (g)  F(g)       DE = +77 kJ
                   E to break F-F bond

45: F(g) + 1e-  F-1      DE = -328 kJ
                   Electron Affinity             DEnet = -570 kJ/mol

56: Na+(g) + F-(g)  NaF(g) DE = -923 kJ
                   Lattice Energy
Figure 9.6   The Born-Haber cycle for lithium fluoride
       Which is bigger?
The formation of MgO or NaF?
                    Why is Oxygen’s
                    second electron affinity
                    endothermic?
                   MgO
Mg  Mg2+ + 2e-    DH = 2188 kJ

O + 1e-  O-1      DH = -141 kJ
O-1 + 1e-  O-2        DH = 878 kJ
O + 2e-  O-2      DH = 737 kJ

    Why does MgO form????????

Mg2+ + O-2  MgO         DH = 3923 kJ/mol
               Lattice Energy
The change in energy that takes place when
  separated gaseous ions are packed together
  to form an ionic solid

Ex: M+ (g) + X- (g)  MX(l) + Energy
                    (Q1) (Q2)
        LE = k
                          r
   Q is charge of ion
   R is distance b/w ions
   LE depends on charge more than radius/size
                               Melting Point (C)




                                250
                                500
                                750
                               1000
                               1250
                               1500
                               1750
                               2000
                               2250
                               2500
                               2750
                               3000




                                  0
                    Zn
                        C
                    Fe l2
                        C
                   Na l3
                      NO
                    KN 3
                        O
                    Cu 3
                        B
                    Cu r 2
                        C
                    M l2
                      nC
                    M l2
                      gC
                    Co l2
                        Cl
                           2
                      KB
                           r
                     Na
                         Br
                       KC
                    Ca l
                        Cl
                           2
                     Na
                   Ni Cl
                      S
                  Na O4
                    2S
Ionic Compounds   K2 O4
                     Cr
                        O
                           4
                     Ni
                        C
                          l2
                      Cu
                   Fe O
                      2O
                           3
                                                   Melting Point for Ionic Compounds




                      Ca
                          O
                     M
                        gO
Sizes of Ions On P.T.
As size of ions increase,
Lattice energy ________




As size of charge increase,
Lattice energy ________
          Sample Problem
Which compound in each of the following pairs
 of ionic substances has the most exothermic
      lattice energy? Justify your answers.

              NaCl vs. KCl

              MgO vs. NaCl

              Fe(OH)2 vs Fe(OH)3
               NaCl vs. KCl
The larger the ionic
  radius, the lower the
  lattice energy

The amount of
  interaction between
  the ions is smaller
  and the packing of the
                           L.E. NaCl = -780 kJ/mol
  ions is less efficient   L.E. KCl = -711 kJ/mol
               MgO vs. NaCl
The stronger the charge
  on an ion the stronger
  the attractive force
  that will result in an
  ionic lattice.

+/-1 ions form
  compounds with
  lower lattice energies
                              L.E. MgO = -3791 kJ/mol
  than +/-2 ions.             L. E. NaCl = -780 kJ/mol
Fe(OH)2 vs. Fe(OH)3
Figure 9.9            Electrical Conductance and Ion Mobility




             Solid ionic         Molten ionic       Ionic compound
             compound            compound          dissolved in water
     Summary of Ionic Bonds
Between metal ion and non – metal ion
Electrostatic attraction between ions
Veeerrrrrryyyyy strong!
             Covalent Bond
• Sharing of electron pairs between two
  atoms
• Shared electrons are simultaneously
  attracted to both atomic nuclei
Relationship b/w Bond Length & Bond Energy
 Graph: Chang Disc 2


                                                  a
                              d       b

                                  c




a) Large distances away; no influence on each other; Ep = 0
b) Atoms move closer together, nuclei are attracted to each other;Ep ↓
c) Attractive force dominates; Ep is at a minimum STABLE!
d) If nuclei move closer together, repulsive force > attractive force; Ep ↑
  Lone Pairs vs. Bonding Pairs
Lone pairs:
  electron pairs not involved in bonding

Bonding Pairs:
 electron pairs involved in bonding
             Bond Length
Distance between two atomic nuclei when
  energy is at its minimum (when they are
  the most stable)
Figure 9.13          Bond length and covalent radius.
      Internuclear distance   Covalent   Internuclear distance    Covalent
          (bond length)        radius        (bond length)         radius




     Internuclear distance    Covalent    Internuclear distance   Covalent
         (bond length)         radius         (bond length)        radius
Relationship between Bond Energy
         and Bond Length

                                      Bond Energy vs. Bond Length
                        500
                        450
                        400
 Bond Energy (kJ/mol)




                        350
                        300
                        250
                        200
                        150       BE = -1.6437kJ/mol/pm)BL + 587.6 kJ/mol
                        100                     R2 = 0.9087
                         50
                          0
                              0          50       100         150           200   250   300
                                                        Bond Length (pm )
SAMPLE PROBLEM 9.2           Comparing Bond Length and Bond Strength


PROBLEM:       Using the periodic table, but not Tables 9.2 and 9.3, rank the
               bonds in each set in order of decreasing bond length and bond
               strength:
               (a) S - F, S - Br, S - Cl     (b) C = O, C - O, C   O
PLAN:      (a) The bond order is one for all and sulfur is bonded to halogens;
           bond length should increase and bond strength should decrease
           with increasing atomic radius. (b) The same two atoms are
           bonded but the bond order changes; bond length decreases as
           bond order increases while bond strength increases as bond order
           increases.
SOLUTION:
(a) Atomic size increases                    (b) Using bond orders we get
going down a group.
Bond length: S - Br > S - Cl > S - F       Bond length: C - O > C = O > C        O

Bond strength: S - F > S - Cl > S - Br     Bond strength: C   O>C=O>C-O
  Strong forces within molecules and weak forces between them.


Figure 9.14                  Strong covalent bonding forces within molecules




                            Weak intermolecular forces between molecules
Figure 9.15   Covalent bonds of network covalent solids.
     Covalent Bond Energies &
       Chemical Reactions
Recall relationship between bond length and
 bond energy…
  Bond Energy – Average of
   Individual Bond Energies

Bond energy depends on the environment

            C—H Bond Energy (kJ/mol)
   i.e. HCBr3     380
        HCCl3     380
        HCF3      430
        C2H6      410

       Table 8.4   413
     Types of Covalent Bonds
single bond— 1 e- pair shared

double bond— 2 e- pairs shared

triple bond— 3 e- pairs shared
Bond Type
                 Demo:
Testing for double bonds (Br2)
         Demo: Br2 water

Hexane       Cyclohexene




                           Vegetable Oil
   Bond Energy and Enthalpy
What kind of energy changes accompany
 this reaction?

    2H2 (g) + O2 (g)  2 H2O (g)




     break these bonds
                         form these bonds
 2 H2 (g)      +      O2 (g)        2 H2O(g)



     break these bonds               form these bonds
                     O O HHHH
               O—O
               H H
               H H
Energy
         O—O            H—O—H
         H—H            H—O—H   DH
         H—H
                     Equation
2H2 (g)      +      O2 (g)              2 H2O (g)

DHorxn=SBEreactant bonds broken–SBEproduct bonds formed

           Energy Required             Energy Released


Where BE = bond energy per mole of bonds
              Sample Problem
Calculate the DH for the reaction of methane
 with chlorine and fluorine to give Freon – 12

CH4 (g)+2 Cl2 (g)+ 2 F2 (g)CF2Cl2 (g) + 2 HF(g) +2 HCl(g)




Note: Calculating DH = SDHfo (products) – SDHfo (reactants)
                     is more accurate
           Electronegativity
The ability of an atom in a molecule to
 attract shared electrons to itself
            Pauling Electronegativity Values
 How To Determine Bond Type
Calculate DEN (Absolute Value)
The bigger the DEN, the (more, less) the
 electrons are shared
            Pauling Electronegativity Values
Figure 9.16   The Pauling electronegativity (EN) scale.
Figure 9.17   Electronegativity and atomic size.
 SAMPLE PROBLEM 9.3          Determining Bond Polarity from EN Values

PROBLEM:     (a) Use a polar arrow to indicate the polarity of each bond:
             N-H, F-N, I-Cl.
             (b) Rank the following bonds in order of increasing polarity:
             H-N, H-O, H-C.

PLAN:     (a) Use Figure 9.16(button at right) to find EN values; the
          arrow should point toward the negative end.
          (b) Polarity increases across a period.

SOLUTION: (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0

               N-H                     F-N                    I - Cl

             (b) The order of increasing EN is C < N < O; all have an EN
             larger than that of H.
                            H-C < H-N < H-O
                                    3.0




 Figure 9.18                        2.0
                              DEN

Boundary ranges for
classifying ionic character
of chemical bonds.



                                    0.0
Figure 9.19
         Percent ionic character of electronegativity difference (DEN).
Figure 9.20




              Li   F
             Electrons                % Ionic
                           DEN                   Electron Cloud
               are…                  Character


 Covalent
              shared       0-0.3      <5%
(Non-Polar
              equally
 Covalent)


 Polar        shared       0.3-1.6     5-49%
Covalent      unequally

             transferred
  Ionic      from          >1.7        >50%
             metal to
             non-metal
Difference in Electronegativity
                                  3.3                         100 %




                                                                      Percent Ionic Character
                                               Ionic

                                  1.7                         50 %


                                         Polar - Covalent
                                  0.3                         5%

                                    0   Nonpolar - covalent   0%
                Demo!
Compare the conductivity of an aqueous
 ionic compound to an aqueous covalent
 compound
       Bond Polarity Notation
A bond or molecule that is dipolar (polar) is
  said to have a dipole moment
    Structure of a metallic crystal




The “sea” of delocalized electrons allows atoms to move past one
another as the solid deforms, creating a malleable substance.
           Properties of Metals
• Conduct heat and electricity (in2A(2) elements.
   Melting points of the Group 1A(1) and Group
                                               solid and
  liquid phases)
• Malleable
• Ductile
• High Melting and Boiling Points
Numisticists Love Mallebility
  Crystalline Solids often have
             defects




Atomic defects distort the crystal and
make the motion of atoms more difficult
   VitreloyTM—an amorphous alloy
The random
arrangement of the
different sized atoms
prevents them from
moving past each
other.

          Zr41.2Be22.5 Ti13.8Cu12.5Ni10.0
    The Importance of Structure!




Different size atoms create a solid that behaves as a large
collection of defects with no long range crystal order. The
resulting malleability is greatly decreased. (see video clip)
           Examine The Evidence




Small pits in the stainless steel
base are evidence of the            At 7.5x
permanent deformation that          magnification,
occurs when the ball bearing is     the pitting of
dropped onto the surface. This      the stainless
deformation accounts for much       steel base is
of the energy dissipation of the
                                    clear.
bouncing ball.
 Comparison of steel vs amorphous
              metal




Steel deformation > 7.5 x Amorphous metal deformation
Liquid Metal

								
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